Screening Test: Base Rates and Confirmatory Testing
A school district screens incoming students for a specific learning difference. The screening instrument has sensitivity 92% and specificity 88%, and district data indicate that about 6% of incoming students actually have the difference. Results are reported as positive or negative. The vendor's brochure states, 'With 92% accuracy, a positive result means the student very likely has the difference.' A parent whose child screened positive asks the district psychologist what the result means and whether retesting with the same instrument would help. The psychologist decides to explain the result using a hypothetical cohort of 10,000 students and to check whether a second, independent positive result would change the interpretation.
1. Using a cohort of 10,000 students, compute the probability that a student who screens positive actually has the difference.
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Build the cohort.
- Affected: 6% of 10,000 = 600.
- Unaffected: 9,400.
Apply the test.
- True positives: 0.92 × 600 = 552.
- False negatives: 600 − 552 = 48.
- False positives: (1 − 0.88) × 9,400 = 0.12 × 9,400 = 1,128.
- True negatives: 9,400 − 1,128 = 8,272.
Positive predictive value. P(affected | positive) = 552 ÷ (552 + 1,128) = 552 ÷ 1,680 = 0.3286.
So only about 32.9% - roughly one in three - of positive screens are true positives. The base rate of 6% dominates the result.
2. Evaluate the brochure's claim. State what the 92% figure actually measures, and compute the test's overall accuracy for comparison.
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What 92% is. It is the sensitivity, P(positive | affected). The brochure treats it as P(affected | positive), which is the reverse conditional; these are equal only when prevalence is 50%, which is not the case here.
Overall accuracy. (552 + 8,272) ÷ 10,000 = 8,824 ÷ 10,000 = 88.2%. Even this figure is inflated as a guide to a positive result, because the 1,128 false positives outnumber the 552 true positives by about 2.04 to 1.
Conclusion. The claim 'a positive result means the student very likely has the difference' is false. A positive screen raises the probability from 6% to about 33%, which is meaningful but far from diagnostic. The counselor should describe the result as a flag for further evaluation, not as a finding.
3. A positive student is retested with an independent instrument having the same sensitivity and specificity, and the second result is also positive. Compute the updated probability and judge whether the district's 'confirm with a second screen' policy is sound.
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Update. Use the first posterior, 0.3286, as the prior for the second test.
P(affected | two positives) = (0.3286)(0.92) / [(0.3286)(0.92) + (0.6714)(0.12)] = 0.3023 / (0.3023 + 0.0806) = 0.3023 / 0.3829 ≈ 0.790.
Interpretation. Two positive screens raise the probability to about 79%, a large improvement over 33%. The policy is directionally sound, but about 21% of twice-positive students would still not have the difference, so a second screen cannot replace a full psychoeducational evaluation.
Caveat. The calculation assumes the two results are conditionally independent given status. If the same instrument, rater, or conditions are reused, the errors correlate and the true posterior is lower; 'independent' must be enforced, not assumed.